I've caught by a loop of:

Standard texts of Non-Relativistic Quantum Mechanics $\to$ Representation theory of Lie groups and Lie algebras of $SO(3)$ and $SU(2)$ $\to$ Discussions of infinitesimal transformations and formulas like $\Lambda_{\alpha}^{\beta} = \mathrm{exp}\Big(\frac{1}{2}\Omega_{\mu\nu}(M^{\mu\nu})_{\alpha}^{\beta}\Big)$ and $S[\Lambda]_{a}^{b} = \mathrm{exp}\Big(\frac{1}{2}\Omega_{\mu\nu}(S^{\mu\nu})_{a}^{b}\Big)$ $\to$ Classical Field theory and Noether theorem $\to$ Active/Passive trasformations $\to$ Standard texts of Relativistic Quantum Mechanics

But what I really want to understand is simply: what are Pauli spinors! For tensors is simple: they are elements of the tensor product. But for some reason the active/passive views ARE important.

Now, to be clear in this question, consider the Group: $SO(1,3)^{+}_{\uparrow}$. I understand completely when someone writes:

$$V^{\mu'} = \Lambda^{\mu'}_{\nu}V^{\nu}, \tag{1}$$

this is a 4-vector. I don't even mentioned the words "representation", "realization", "fundamental representation" etc... I use just the common sense notion: "hey, I want to transform this vector field in spacetime and I use these matrices because of Lorentz symmetry".

But, I do not understand when someone writes:

$$\psi' = S[\Lambda] \psi, \tag{2}$$

and call this a spinor! Where are its indices?

Now, to elaborate more:

a wavefunction is a scalar field, period. A pauli spinor and generally a Dirac spinor cannot be scalar fields, and therefore cannot be simple wave functions. So my question is:

since from non-relativistic quantum mechanics a Pauli spinor is: $\big(\langle \vec{r}| \otimes \langle s|\big) |\psi\rangle := \psi(\vec{r},s) \equiv \begin{pmatrix}\psi(\vec{r},\uparrow) \\ \psi(\vec{r},\downarrow)\end{pmatrix} \equiv \begin{pmatrix}\psi(\vec{r},1) \\ \psi(\vec{r},2)\end{pmatrix} := \psi^{a}(\vec{r})$ with $a={1,2}$, the index $a$ is from the lie algebra or from the spin? In other words, are the lie algebra indices synomys to spinor indices?

  • $\begingroup$ Are the quote blocks from a reference? Which pages? $\endgroup$
    – Qmechanic
    Apr 7, 2023 at 10:20
  • $\begingroup$ The first one is just to explain how I'm struggle with spinors. The second one is from "Quantum Mechanics - A.F.R de Toledo Piza, page 469" unfortunately is available just in portuguese. $\endgroup$ Apr 7, 2023 at 10:24

1 Answer 1


Let's take a step back: There are two ways to do linear algebra - in components and abstractly. For various reasons, physics texts often only do it in components (and this is the reason the question "where are its indices?" appears here).

When you have a vector space $V$ and a linear operator $A : V\to V$ on that vector space, then you can either write $Av$ to denote the action of $A$ on a vector $v\in V$, or you can choose a basis $e_i \in V$ and write the action of $A$ on a vector $v$ in terms of the matrix components as $(Av)^i = {A^i}_j v^j$.

Eq. (1) in the question is written in components, eq. (2) is not. They are equally statements about the action of certain linear operators on a vector space.

When we talk about non-relativistic spinors, we're talking about certain representations of the rotation algebra $\mathfrak{su}(2)$, i.e. we have some vector spaces $V_\rho$ with representation maps $\rho : \mathfrak{su}(2)\to \mathfrak{gl}(V_\rho)$, where $\mathfrak{gl}(V_\rho)$ is just the algebra of all linear operators on $V_\rho$.

For any such representation, the action of an infinitesimal rotation $\omega\in\mathfrak{su}(2)$ on a vector $v\in V_\rho$ can be written as either $$ v' = \rho(\omega) v \tag{abstract}$$ or $$ v'^i = {\rho(\omega)^i}_j v^j, \tag{components}$$ regardless of what representation it is - the indices come from a basis of $V_\rho$ and run from $1$ to $\dim(V_\rho)$.

It is customary in physics to use Greek indices when $V_\rho$ is Minkowski space, Latin indices starting from $i$ when $V_\rho$ is Euclidean space, and Latin indices starting from $a$ when $V_\rho$ is something else.

Non-relativistic spinors have a two-dimensional $V_\rho$, and so this falls into the "something else" category and the component notation will denote these spinors with $\psi^a$ in this convention.

Crucially, you have to distinguish this from an index on the Lie algebra $\mathfrak{su}(2)$ - as a Lie algebra, this is also a vector space, and you will see people talking about a basis $T^a$ of this vector space. Just because both indices are denoted with $a$ in these different contexts does not mean that this index is "the same" as that on the spinor $\psi^a$. The spinor index runs from 1 to 2, the index on the generators $T^a$ from 1 to 3.

  • $\begingroup$ So in Pauli equation, $H\psi = i\partial_{t}\psi$, how can I write it using indices? And where lies the lie algebra indices, since the $a$ will be the spinor ones? $\endgroup$ Apr 7, 2023 at 17:10
  • $\begingroup$ Because, my reasoning with spinor indices tends to write $\{\frac{1}{2m}[(-i\nabla_{i}-eA_{j})^2-e\sigma^{j}_{ab}B_{j}]+\delta_{ab}e\phi\}\psi^{a} = i\partial_{t}\psi^{b}$. But where are the lie algebra indices? $\endgroup$ Apr 7, 2023 at 17:13
  • $\begingroup$ And the reason why we use $SU(2)$, a mathematical argument at least, is because $SO(3)$ doesn't have spinor representation? Therefore we are forced to represent it's double cover SU(2)? $\endgroup$ Apr 7, 2023 at 17:21
  • $\begingroup$ @BasicMathGuy 1. I'm not 100% sure what you mean by a "Lie algebra index" - what do you think the $j$ on your $\sigma^j_{ab}$ is? Why do you think there should be "Lie algebra indices" in this equation? 2. See this answer of mine for why the universal cover appears in quantum mechanics. $\endgroup$
    – ACuriousMind
    Apr 7, 2023 at 17:35
  • $\begingroup$ Hey, I guess you're write....$j$ runs from $1,2,3$ which generates the whole thing. But It brings a confunsion for me too, since in gamma matrices $\gamma^{\mu}_{ab}$ the $\mu$ are the spacetime indices, but on its non-relativistic counterpart seems to be the $j$ in $\sigma^{j}_{ab}$ which are not spatial indices. $\endgroup$ Apr 7, 2023 at 17:58

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